Polynomial and Rational Functions

Polynomial Degree and End Behavior

When you graph a polynomial, the middle of the graph can have turns, dips, and crossings. But far to the left and far to the right, the graph follows a simpler pattern. This pattern is called end behavior.

Understanding end behavior helps you quickly predict what a graph does as the input values become very large positive or very large negative numbers. In Grade 12, this is an important skill for sketching polynomial graphs and connecting equations to their shapes.

To predict end behavior, the two most important ideas are:

• the degree of the polynomial, and
• the leading coefficient.

Once you know these, you can use the leading coefficient test to decide how the graph behaves at both ends.

1. What is the degree of a polynomial?

The degree of a polynomial is the highest exponent of the variable when the polynomial is written in standard form.

For example:

• \(f(x) = 3x^4 - 2x + 7\) has degree \(4\)
• \(g(x) = -5x^3 + x^2 - 9\) has degree \(3\)
• \(h(x) = 6x - 1\) has degree \(1\)

The term with the highest exponent is called the leading term, and its coefficient is called the leading coefficient.

So for \(f(x) = 3x^4 - 2x + 7\):

• leading term: \(3x^4\)
• leading coefficient: \(3\)

2. Why does the leading term control end behavior?

As \(x\) becomes very large or very negative, the highest power of \(x\) grows much faster than the lower-degree terms. That means the leading term has the biggest effect on the value of the polynomial.

For example, in

$$f(x) = 2x^5 - 4x^2 + 1$$

when \(x\) is very large, the term \(2x^5\) dominates the terms \(-4x^2\) and \(1\). So the end behavior of \(f(x)\) is the same as the end behavior of \(2x^5\).

This is why, for end behavior, you usually only need to look at the leading term.

3. Even degree vs odd degree

The parity of the degree means whether the degree is even or odd.

• Even degree: \(2, 4, 6, 8, \dots\)
• Odd degree: \(1, 3, 5, 7, \dots\)

This matters because even powers and odd powers behave differently when \(x\) is negative.

Even degree:

• For large positive \(x\), \(x^n\) is positive.
• For large negative \(x\), \(x^n\) is also positive if \(n\) is even.

Odd degree:

• For large positive \(x\), \(x^n\) is positive.
• For large negative \(x\), \(x^n\) is negative if \(n\) is odd.

4. The leading coefficient test

The leading coefficient test uses the degree and the sign of the leading coefficient to predict the end behavior of a polynomial function.

Suppose

$$f(x) = a_nx^n + \text{lower-degree terms}$$

where \(a_n\) is the leading coefficient and \(n\) is the degree.

Then the end behavior depends on two things:

• whether \(n\) is even or odd,
• whether \(a_n\) is positive or negative.

Case 1: Even degree, positive leading coefficient

The graph rises on both ends.

$$\text{As } x \to -\infty,\ f(x) \to \infty$$

$$\text{As } x \to \infty,\ f(x) \to \infty$$

Case 2: Even degree, negative leading coefficient

The graph falls on both ends.

$$\text{As } x \to -\infty,\ f(x) \to -\infty$$

$$\text{As } x \to \infty,\ f(x) \to -\infty$$

Case 3: Odd degree, positive leading coefficient

The graph falls to the left and rises to the right.

$$\text{As } x \to -\infty,\ f(x) \to -\infty$$

$$\text{As } x \to \infty,\ f(x) \to \infty$$

Case 4: Odd degree, negative leading coefficient

The graph rises to the left and falls to the right.

$$\text{As } x \to -\infty,\ f(x) \to \infty$$

$$\text{As } x \to \infty,\ f(x) \to -\infty$$

5. A quick memory guide

• Even degree means both ends go the same way.
• Odd degree means the ends go opposite ways.
• Positive leading coefficient means the graph goes up on the right.
• Negative leading coefficient means the graph goes down on the right.

If you remember just one thing, remember this: look at the right end first. The sign of the leading coefficient tells you whether the graph rises or falls on the right. Then use even/odd degree to decide what happens on the left.

6. Worked Example 1

Determine the end behavior of

$$f(x) = 4x^3 - 2x + 7$$

Step 1: Find the leading term.

The leading term is \(4x^3\).

Step 2: Identify the degree and leading coefficient.

• Degree: \(3\), which is odd
• Leading coefficient: \(4\), which is positive

Step 3: Use the leading coefficient test.

Odd degree and positive leading coefficient means:

• left end down
• right end up

So,

$$\text{As } x \to -\infty,\ f(x) \to -\infty$$

$$\text{As } x \to \infty,\ f(x) \to \infty$$

Interpretation: The graph falls to the left and rises to the right.

7. Worked Example 2

Determine the end behavior of

$$g(x) = -3x^4 + 5x^2 - 1$$

Step 1: Find the leading term.

The leading term is \(-3x^4\).

Step 2: Identify the degree and leading coefficient.

• Degree: \(4\), which is even
• Leading coefficient: \(-3\), which is negative

Step 3: Use the leading coefficient test.

Even degree and negative leading coefficient means both ends go down.

So,

$$\text{As } x \to -\infty,\ g(x) \to -\infty$$

$$\text{As } x \to \infty,\ g(x) \to -\infty$$

Interpretation: The graph falls on both the left and right ends.

8. Worked Example 3

Determine the end behavior of

$$h(x) = -2x^5 + x^4 + 7x - 9$$

This example is useful because the polynomial has several terms, but only one determines the end behavior.

Step 1: Find the leading term.

The leading term is \(-2x^5\).

Step 2: Identify the degree and leading coefficient.

• Degree: \(5\), which is odd
• Leading coefficient: \(-2\), which is negative

Step 3: Use the leading coefficient test.

Odd degree and negative leading coefficient means:

• left end up
• right end down

So,

$$\text{As } x \to -\infty,\ h(x) \to \infty$$

$$\text{As } x \to \infty,\ h(x) \to -\infty$$

Interpretation: The graph rises to the left and falls to the right.

9. Worked Example 4

A polynomial has degree \(6\) and leading coefficient \(7\). Predict its end behavior without knowing the full equation.

Step 1: Identify the type.

• Degree \(6\) is even
• Leading coefficient \(7\) is positive

Step 2: Apply the rule.

Even degree and positive leading coefficient means both ends rise.

Therefore,

$$\text{As } x \to -\infty,\ f(x) \to \infty$$

$$\text{As } x \to \infty,\ f(x) \to \infty$$

This shows that you do not always need the whole polynomial. Sometimes degree and leading coefficient are enough.

10. How this helps when sketching graphs

When sketching a polynomial graph, end behavior gives you the overall direction of the graph at the far left and far right. Then you can combine that with other information such as:

• real roots or x-intercepts,
• whether roots repeat,
• turning points,
• the y-intercept.

For example, if a polynomial has odd degree and a positive leading coefficient, you already know the graph must start low on the left and end high on the right. That makes sketching much more accurate.

11. Common mistakes to avoid

• Looking at the constant term instead of the leading term.
  The constant term affects the graph near the y-axis, not the end behavior.
• Forgetting whether the degree is even or odd.
  Even means same direction on both ends. Odd means opposite directions.
• Ignoring the sign of the leading coefficient.
  The sign decides whether the right end goes up or down.
• Using all the terms instead of focusing on the highest-degree term.
  For end behavior, the leading term is the key.

12. Quick check questions

1. For \(f(x) = x^8 - 3x + 1\), what is the end behavior?
   Answer: Even degree, positive leading coefficient, so both ends rise.
2. For \(g(x) = -x^3 + 2x^2 - 4\), what is the end behavior?
   Answer: Odd degree, negative leading coefficient, so left rises and right falls.
3. For \(h(x) = -6x^2 + 5\), what is the end behavior?
   Answer: Even degree, negative leading coefficient, so both ends fall.

13. Summary

To determine the end behavior of a polynomial, look at its leading term. The degree tells you whether the ends go in the same direction or opposite directions, and the leading coefficient tells you whether the right end goes up or down.

Use the leading coefficient test:

• Even degree, positive coefficient: up on both ends
• Even degree, negative coefficient: down on both ends
• Odd degree, positive coefficient: down left, up right
• Odd degree, negative coefficient: up left, down right

Once you master this, you will be able to predict the long-run shape of polynomial graphs quickly and confidently.

Polynomial Division Methods are essential tools for working with polynomial and rational functions. In Grade 12, you use them to simplify expressions, factor polynomials, find zeros, and determine features of graphs such as slant asymptotes.

When one polynomial is divided by another, the result may be another polynomial, sometimes with a remainder. This is similar to ordinary number division, but now the “numbers” are algebraic expressions.

In this lesson, you will learn two main methods:

• Polynomial long division
• Synthetic division

You will also see how these methods help with factoring polynomials and finding slant asymptotes of rational functions.

1. The division idea

If a polynomial \(P(x)\) is divided by a polynomial \(D(x)\), then the result can be written as

where:

• \(P(x)\) is the dividend
• \(D(x)\) is the divisor
• \(Q(x)\) is the quotient
• \(R(x)\) is the remainder

The degree of the remainder must be less than the degree of the divisor.

For example, if you divide a cubic polynomial by a linear polynomial, the quotient will usually be quadratic, and the remainder will be a constant.

2. Polynomial long division

Polynomial long division works much like regular long division with numbers. It is especially useful when the divisor is not in the form \(x-c\), or when synthetic division cannot be used easily.

Steps for polynomial long division

1. Write both polynomials in descending powers of \(x\).
2. Insert any missing terms with coefficient 0.
3. Divide the first term of the dividend by the first term of the divisor.
4. Write that result in the quotient.
5. Multiply the entire divisor by that quotient term.
6. Subtract.
7. Bring down the next term and repeat until the remainder has smaller degree than the divisor.

Worked Example 1: Long division with no remainder

Divide \(x^3+5x^2+6x+8\) by \(x+2\).

Step 1: Divide the first terms.

Put \(x^2\) in the quotient.

Step 2: Multiply \(x^2(x+2)=x^3+2x^2\).

Subtract:

Step 3: Divide the new first term.

Put \(+3x\) in the quotient.

Multiply:

Subtract:

Step 4: Divide again.

This is not a polynomial term, and the remainder has degree less than the divisor, so we stop.

So the result is

or, in division form,

Important note: Since the remainder is 8, \(x+2\) is not a factor of the polynomial.

3. Synthetic division

Synthetic division is a shortcut method for dividing a polynomial by a linear divisor of the form

It is faster than long division, but it only works for divisors that are linear and have leading coefficient 1.

How synthetic division works

1. Write the coefficients of the polynomial in order.
2. Use the value \(c\) from \(x-c\).
3. Bring down the first coefficient.
4. Multiply by \(c\), then add to the next coefficient.
5. Repeat until all coefficients are used.
6. The final number is the remainder.

Worked Example 2: Synthetic division with remainder 0

Divide \(2x^3-3x^2-11x+6\) by \(x-3\).

Since the divisor is \(x-3\), use \(c=3\).

Coefficients: \(2, -3, -11, 6\)

So the quotient is

and the remainder is \(0\).

This means

Now factor the quadratic:

So the full factorization is

From this, the roots are

Why this matters: Division helps turn a large polynomial into smaller factors, which makes finding roots much easier.

4. The Remainder Theorem and Factor Theorem

These theorems connect division and roots.

Remainder Theorem: If a polynomial \(P(x)\) is divided by \(x-c\), then the remainder is \(P(c)\).

This means synthetic division can also be used to evaluate a polynomial quickly.

Factor Theorem: \((x-c)\) is a factor of \(P(x)\) if and only if \(P(c)=0\).

So if the remainder is zero when dividing by \(x-c\), then \(x-c\) is a factor, and \(c\) is a root.

Worked Example 3: Using synthetic division to factor completely

Factor \(x^3-4x^2-x+4\).

We can test possible rational roots such as \(1, -1, 2, -2, 4, -4\).

Try \(x=1\):

The remainder is 0, so \((x-1)\) is a factor.

The quotient is

Now factor the quadratic:

So the full factorization is

The roots are

This factorization is useful for graphing because the roots give the x-intercepts of the polynomial.

5. Missing terms must be included

One common mistake in both long division and synthetic division is forgetting missing powers of \(x\).

For example, in \(x^4-5x+2\), the missing terms are \(0x^3\) and \(0x^2\).

You must write it as

This keeps the coefficients in the correct positions.

6. Long division and rational functions

Polynomial division is also important when studying rational functions. A rational function is a quotient of two polynomials, such as

If the degree of the numerator is greater than or equal to the degree of the denominator, division helps rewrite the function in a form that is easier to understand and graph.

7. Slant asymptotes

A slant asymptote happens when:

• the degree of the numerator is exactly one more than the degree of the denominator

To find the slant asymptote, divide the numerator by the denominator. The quotient, without the remainder part, gives the asymptote.

Worked Example 4: Finding a slant asymptote

Find the slant asymptote of

Use long division.

Step 1: Divide the first terms.

Put \(x\) in the quotient.

Multiply:

Subtract:

Step 2: Divide again.

Put \(+3\) in the quotient.

Multiply:

Subtract:

So

As \(x\) becomes very large positive or negative, the term

gets very close to 0.

Therefore, the graph approaches

So the slant asymptote is

8. When to use each method

Use synthetic division when:

• the divisor is linear
• the divisor has the form \(x-c\)
• you want a faster method
• you are testing possible roots

Use polynomial long division when:

• the divisor is not in the form \(x-c\)
• the leading coefficient of the divisor is not 1
• you are finding slant asymptotes
• you want a method that always works

9. Common mistakes to avoid

• Using the wrong sign in synthetic division. If dividing by \(x-2\), use \(2\). If dividing by \(x+2\), use \(-2\).
• Forgetting missing terms. Always include zero coefficients.
• Stopping too early in long division. Continue until the remainder has lower degree than the divisor.
• Confusing quotient and remainder. The last number in synthetic division is the remainder.
• Writing the asymptote incorrectly. For a slant asymptote, use only the quotient, not the remainder term.

10. Quick check of understanding

Ask yourself:

• Can I divide a polynomial by \(x-c\) using synthetic division?
• Can I tell whether a remainder of 0 means there is a factor?
• Can I use long division to rewrite a rational function?
• Can I identify a slant asymptote from the quotient?

Summary

Polynomial long division and synthetic division are key methods for simplifying and analyzing polynomial expressions. Synthetic division is a fast shortcut for divisors of the form \(x-c\), while long division works for all polynomial divisors.

These methods help you factor polynomials, find roots, use the Remainder and Factor Theorems, and determine slant asymptotes of rational functions. Mastering these skills makes graphing polynomial and rational functions much easier.

Remainder and Factor Theorems are powerful tools for working with polynomials. They help us evaluate polynomial values quickly and decide whether a binomial such as \((x-a)\) is a factor of a polynomial.

These ideas are especially useful when finding roots, checking factors, and understanding the structure of polynomial functions. They also connect directly to graphing, because factors help identify the x-intercepts of a polynomial.

In this lesson, you will learn:

• what the Remainder Theorem says,
• what the Factor Theorem says,
• how to use substitution to evaluate remainders,
• how to test whether a binomial is a factor,
• and how these ideas help with roots and graphs.

1. A quick review: division of polynomials

When one polynomial is divided by another, the result can be written in the form

$$P(x)=D(x)Q(x)+R(x)$$

where:

• \(P(x)\) is the original polynomial,
• \(D(x)\) is the divisor,
• \(Q(x)\) is the quotient,
• and \(R(x)\) is the remainder.

If the divisor is linear, such as \((x-a)\), then the remainder must be a constant number.

For example, if we divide a polynomial by \((x-3)\), the remainder will just be some number like \(5\), \(-2\), or \(0\).

2. The Remainder Theorem

The Remainder Theorem says:

$$\text{If a polynomial } P(x) \text{ is divided by } (x-a), \text{ then the remainder is } P(a).$$

This means you do not need to carry out long division every time. To find the remainder when dividing by \((x-a)\), you can simply substitute \(a\) into the polynomial.

So:

• remainder when dividing by \((x-2)\) is \(P(2)\),
• remainder when dividing by \((x+5)\) is \(P(-5)\),
• remainder when dividing by \((x-\tfrac12)\) is \(P(\tfrac12)\).

Important sign reminder: if the divisor is \((x-a)\), use \(a\). If the divisor is \((x+b)\), that is really \((x-(-b))\), so use \(-b\).

3. Why the Remainder Theorem works

Suppose

$$P(x)=(x-a)Q(x)+r$$

where \(r\) is the remainder. Since we are dividing by a linear factor, the remainder is a constant.

Now substitute \(x=a\):

$$P(a)=(a-a)Q(a)+r$$

$$P(a)=0+r$$

$$P(a)=r$$

So the remainder is exactly the value of the polynomial at \(x=a\).

4. The Factor Theorem

The Factor Theorem is a direct result of the Remainder Theorem. It says:

$$ (x-a) \text{ is a factor of } P(x) \iff P(a)=0 $$

In words:

• If substituting \(a\) into the polynomial gives \(0\), then \((x-a)\) divides the polynomial exactly.
• If \((x-a)\) is a factor, then \(a\) is a root or zero of the polynomial.

This is very important because roots, factors, and x-intercepts are closely connected:

• \(P(a)=0\) means \(a\) is a root or zero,
• which means \((x-a)\) is a factor,
• which means the graph has an x-intercept at \((a,0)\).

5. Worked Example 1: Find a remainder

Find the remainder when

$$P(x)=2x^3-5x^2+4x-7$$

is divided by \((x-2)\).

Step 1: Use the Remainder Theorem.

The remainder is \(P(2)\).

Step 2: Substitute \(x=2\).

$$P(2)=2(2^3)-5(2^2)+4(2)-7$$

$$=2(8)-5(4)+8-7$$

$$=16-20+8-7$$

$$=-3$$

Answer: The remainder is \(-3\).

This is much faster than doing polynomial long division.

6. Worked Example 2: Check whether a binomial is a factor

Determine whether \((x+1)\) is a factor of

$$P(x)=x^3-4x^2+x+6$$

Step 1: Rewrite the divisor in the form \((x-a)\).

Since \((x+1)=(x-(-1))\), we use \(a=-1\).

Step 2: Evaluate \(P(-1)\).

$$P(-1)=(-1)^3-4(-1)^2+(-1)+6$$

$$=-1-4(1)-1+6$$

$$=-1-4-1+6$$

$$=0$$

Step 3: Apply the Factor Theorem.

Since \(P(-1)=0\), the remainder is \(0\).

Answer: \((x+1)\) is a factor of the polynomial.

This also tells us that \(x=-1\) is a root, so the graph crosses or touches the x-axis at \((-1,0)\).

7. Worked Example 3: Find an unknown value using the Factor Theorem

If \((x-3)\) is a factor of

$$P(x)=x^3+kx^2-10x+24,$$

find the value of \(k\).

Step 1: Since \((x-3)\) is a factor, use the Factor Theorem.

$$P(3)=0$$

Step 2: Substitute \(x=3\).

$$P(3)=3^3+k(3^2)-10(3)+24$$

$$=27+9k-30+24$$

$$=21+9k$$

Step 3: Set equal to zero and solve.

$$21+9k=0$$

$$9k=-21$$

$$k=-\frac{21}{9}=-\frac73$$

Answer: $$k=-\frac73$$

This type of question is common because the Factor Theorem turns a factor condition into an equation.

8. Worked Example 4: Use a known factor to help find roots

Given

$$P(x)=x^3-2x^2-5x+6,$$

show that \((x-1)\) is a factor, then find all roots.

Step 1: Test \((x-1)\) using the Factor Theorem.

$$P(1)=1^3-2(1)^2-5(1)+6$$

$$=1-2-5+6$$

$$=0$$

So \((x-1)\) is a factor.

Step 2: Divide the polynomial by \((x-1)\).

Using synthetic division:

$$\begin{array}{r|rrrr} 1 & 1 & -2 & -5 & 6 \\ & & 1 & -1 & -6 \\ \hline & 1 & -1 & -6 & 0 \end{array}$$

So

$$P(x)=(x-1)(x^2-x-6)$$

Step 3: Factor the quadratic.

$$x^2-x-6=(x-3)(x+2)$$

Therefore,

$$P(x)=(x-1)(x-3)(x+2)$$

Answer: The roots are

$$x=1,\quad x=3,\quad x=-2$$

So the graph has x-intercepts at \((1,0)\), \((3,0)\), and \((-2,0)\).

9. Connection to graphing polynomial functions

The Factor Theorem is very useful when sketching polynomial graphs.

If you can factor a polynomial, then you can identify its zeros. Those zeros tell you where the graph meets the x-axis.

For example, if

$$P(x)=(x-2)(x+1)(x-4)$$

then the roots are \(x=2\), \(x=-1\), and \(x=4\). The graph has x-intercepts at

$$ (2,0),\ (-1,0),\ (4,0) $$

So when you test possible factors, you are also testing possible x-intercepts.

10. Common mistakes to avoid

• Using the wrong sign. For divisor \((x+2)\), use \(x=-2\), not \(2\).
• Confusing remainder with quotient. The theorem gives the remainder only.
• Forgetting the Factor Theorem condition. A binomial is a factor only when the remainder is exactly \(0\).
• Arithmetic errors when substituting. Be careful with powers and negative signs.

11. Quick strategy guide

Use this checklist:

1. If asked for the remainder on division by \((x-a)\), compute \(P(a)\).
2. If asked whether \((x-a)\) is a factor, compute \(P(a)\).
3. If \(P(a)=0\), then \((x-a)\) is a factor.
4. If \((x-a)\) is given as a factor and there is an unknown constant, set \(P(a)=0\) and solve.
5. After finding one factor, divide the polynomial to continue factoring and find more roots.

12. Brief summary

The Remainder Theorem says that the remainder when dividing \(P(x)\) by \((x-a)\) is \(P(a)\). The Factor Theorem says that \((x-a)\) is a factor of \(P(x)\) exactly when \(P(a)=0\).

These theorems make polynomial work faster and more meaningful. They help you test factors, find roots, and connect algebra to graphs by identifying x-intercepts of polynomial functions.

Rational Root Theorem and Descartes' Rule

When you are trying to factor a polynomial or find its zeros, it can feel like guessing. Two very helpful tools make the process much more organized: the Rational Root Theorem and Descartes' Rule of Signs.

The Rational Root Theorem gives a finite list of possible rational zeros. Descartes' Rule of Signs helps predict how many positive and negative real zeros a polynomial may have. Together, these tools help you test roots efficiently instead of guessing randomly.

This lesson will show you how each rule works, how to apply them step by step, and how to use them together when factoring polynomials.

1. Review: What is a zero of a polynomial?

A zero or root of a polynomial is a value of \(x\) that makes the polynomial equal to 0.

For example, if

$$f(x)=x^2-5x+6$$

then \(x=2\) and \(x=3\) are zeros because

$$f(2)=0 \quad \text{and} \quad f(3)=0$$

Finding zeros is important because zeros help us:

• factor polynomials,
• solve polynomial equations,
• understand graph behavior,
• find x-intercepts.

2. The Rational Root Theorem

The Rational Root Theorem applies to a polynomial with integer coefficients:

$$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$

If the polynomial has a rational zero of the form

$$\frac{p}{q}$$

in simplest form, then:

• \(p\) must be a factor of the constant term \(a_0\),
• \(q\) must be a factor of the leading coefficient \(a_n\).

So the theorem does not tell you the exact rational roots. It gives all the possible rational roots.

Important idea: after listing the candidates, you must still test them by substitution or synthetic division.

How to generate possible rational zeros

1. Identify the constant term.
2. List all of its factors: these are possible values of \(p\).
3. Identify the leading coefficient.
4. List all of its factors: these are possible values of \(q\).
5. Form all possible fractions \(\pm \frac{p}{q}\).
6. Simplify the list by removing duplicates.

Example: For

$$f(x)=2x^3+3x^2-8x+3$$

The constant term is \(3\), so factors of \(3\) are:

$$\pm 1, \pm 3$$

The leading coefficient is \(2\), so factors of \(2\) are:

$$\pm 1, \pm 2$$

The possible rational zeros are:

$$\pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}$$

3. Descartes' Rule of Signs

Descartes' Rule of Signs helps predict the number of positive and negative real zeros of a polynomial.

Part A: Positive real zeros

Look at the polynomial \(f(x)\) and count the number of sign changes in its coefficients.

A sign change happens when consecutive nonzero coefficients change from positive to negative or from negative to positive.

The number of positive real zeros is equal to:

• the number of sign changes, or
• less than that by an even number.

Part B: Negative real zeros

To find the possible number of negative real zeros, replace \(x\) with \(-x\), making a new polynomial \(f(-x)\). Then count sign changes in \(f(-x)\).

The number of negative real zeros is equal to:

• the number of sign changes in \(f(-x)\), or
• less than that by an even number.

What “less by an even number” means

If there are 3 sign changes, then the possible number is \(3\) or \(1\).

If there are 4 sign changes, then the possible number is \(4\), \(2\), or \(0\).

If there is 1 sign change, then the only possibility is \(1\).

4. Why these tools are useful together

Suppose you have a polynomial of degree 3 or 4 and need to factor it.

The Rational Root Theorem gives a short list of possible rational zeros. Descartes' Rule tells you whether you should expect positive roots, negative roots, or both. This helps you decide which candidates to test first.

For example, if Descartes' Rule says there is exactly 1 positive real root, then positive candidates like \(1\), \(2\), or \(\frac{1}{2}\) are especially worth testing first.

5. Worked Example 1: Listing possible rational zeros

Find all possible rational zeros of

$$f(x)=x^3-4x^2+x+6$$

Step 1: Identify the constant term and leading coefficient

• Constant term: \(6\)
• Leading coefficient: \(1\)

Step 2: List factors

• Factors of \(6\): \(\pm 1, \pm 2, \pm 3, \pm 6\)
• Factors of \(1\): \(\pm 1\)

Step 3: Form \(\pm \frac{p}{q}\)

Since \(q=1\), the possible rational zeros are simply:

$$\pm 1, \pm 2, \pm 3, \pm 6$$

This list does not guarantee all of these are roots. It only says that if a rational root exists, it must be on this list.

6. Worked Example 2: Using Descartes' Rule of Signs

Determine the possible numbers of positive and negative real zeros of

$$f(x)=x^4-3x^3+2x^2+5x-6$$

Positive real zeros

Look at the signs of the coefficients in \(f(x)\):

$$+,-,+,+,-$$

Count sign changes:

• \(+\) to \(-\): 1
• \(-\) to \(+\): 2
• \(+\) to \(+\): no change
• \(+\) to \(-\): 3

So there are 3 sign changes.

The number of positive real zeros is:

$$3 \quad \text{or} \quad 1$$

Negative real zeros

Now compute \(f(-x)\):

$$f(-x)=(-x)^4-3(-x)^3+2(-x)^2+5(-x)-6$$

$$f(-x)=x^4+3x^3+2x^2-5x-6$$

The signs are:

$$+,+,+,-,-$$

Count sign changes:

• \(+\) to \(+\): no change
• \(+\) to \(+\): no change
• \(+\) to \(-\): 1
• \(-\) to \(-\): no change

So there is 1 sign change.

The number of negative real zeros is:

$$1$$

Conclusion:

• Positive real zeros: \(3\) or \(1\)
• Negative real zeros: \(1\)

Since the polynomial has degree 4, any remaining zeros would be non-real complex zeros.

7. Worked Example 3: Using both rules together to factor

Factor

$$f(x)=2x^3-3x^2-8x+12$$

Step 1: Use the Rational Root Theorem

Constant term: \(12\)

Factors of \(12\):

$$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$$

Leading coefficient: \(2\)

Factors of \(2\):

$$\pm 1, \pm 2$$

Possible rational zeros:

$$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}$$

Step 2: Use Descartes' Rule of Signs to guide testing

For \(f(x)=2x^3-3x^2-8x+12\), the signs are:

$$+,-,-,+$$

Sign changes:

• \(+\) to \(-\): 1
• \(-\) to \(-\): no change
• \(-\) to \(+\): 2

So the number of positive real zeros is:

$$2 \quad \text{or} \quad 0$$

Now find \(f(-x)\):

$$f(-x)=2(-x)^3-3(-x)^2-8(-x)+12$$

$$f(-x)=-2x^3-3x^2+8x+12$$

The signs are:

$$-, -, +, +$$

There is 1 sign change, so there is exactly:

$$1 \text{ negative real zero}$$

This suggests trying simple positive and negative candidates.

Step 3: Test possible rational zeros

Try \(x=2\):

$$f(2)=2(2^3)-3(2^2)-8(2)+12=16-12-16+12=0$$

So \(x=2\) is a root, which means \((x-2)\) is a factor.

Use synthetic division:

$$2\;\big|\; 2\quad -3\quad -8\quad 12$$

Bring down the 2:

$$2$$

Multiply and add across:

$$ \begin{array}{r|rrrr} 2 & 2 & -3 & -8 & 12 \\ & & 4 & 2 & -12 \\ \hline & 2 & 1 & -6 & 0 \end{array} $$

So:

$$f(x)=(x-2)(2x^2+x-6)$$

Now factor the quadratic:

$$2x^2+x-6=(2x-3)(x+2)$$

Therefore:

$$f(x)=(x-2)(2x-3)(x+2)$$

Zeros:

$$x=2, \quad x=\frac{3}{2}, \quad x=-2$$

This matches the sign predictions:

• 2 positive real zeros: \(2\) and \(\frac{3}{2}\)
• 1 negative real zero: \(-2\)

8. Worked Example 4: A case where not all possible roots actually work

Analyze

$$f(x)=x^3+x^2-4x-4$$

Step 1: Possible rational zeros

Constant term: \(-4\), so factors are:

$$\pm 1, \pm 2, \pm 4$$

Leading coefficient: \(1\), so the possible rational zeros are:

$$\pm 1, \pm 2, \pm 4$$

Step 2: Descartes' Rule of Signs

For \(f(x)=x^3+x^2-4x-4\), the signs are:

$$+,+,-,-$$

There is 1 sign change, so there is exactly:

$$1 \text{ positive real zero}$$

Now compute \(f(-x)\):

$$f(-x)=(-x)^3+(-x)^2-4(-x)-4=-x^3+x^2+4x-4$$

The signs are:

$$-,+,+,-$$

Sign changes:

• \(-\) to \(+\): 1
• \(+\) to \(+\): no change
• \(+\) to \(-\): 2

So the number of negative real zeros is:

$$2 \quad \text{or} \quad 0$$

Step 3: Test candidates

Try \(x=2\):

$$f(2)=8+4-8-4=0$$

So \((x-2)\) is a factor.

Use synthetic division:

$$ \begin{array}{r|rrrr} 2 & 1 & 1 & -4 & -4 \\ & & 2 & 6 & 4 \\ \hline & 1 & 3 & 2 & 0 \end{array} $$

Thus:

$$f(x)=(x-2)(x^2+3x+2)$$

Factor the quadratic:

$$x^2+3x+2=(x+1)(x+2)$$

So the full factorization is:

$$f(x)=(x-2)(x+1)(x+2)$$

Zeros:

$$x=2, \quad x=-1, \quad x=-2$$

These results fit Descartes' Rule:

• 1 positive real zero
• 2 negative real zeros

9. Common mistakes to avoid

• Forgetting the \(\pm\): If you list factors of the constant term, remember to include both positive and negative possibilities.
• Forgetting to simplify duplicates: For example, \(\frac{2}{2}=1\), so do not list both separately as different options.
• Thinking the Rational Root Theorem gives actual roots: It only gives possible rational roots.
• Miscounting sign changes: Only count when the sign actually changes.
• Making errors in \(f(-x)\): Be careful. Odd powers change sign; even powers do not.
• Ignoring missing terms: If you use synthetic division, make sure every power of \(x\) is represented in the coefficient list, using 0 for missing terms if needed.

10. A step-by-step strategy for solving problems

When you are asked to factor a polynomial or find rational zeros, this method works well:

1. Write the polynomial in standard form.
2. Use the Rational Root Theorem to list all possible rational zeros.
3. Use Descartes' Rule of Signs to predict how many positive and negative real roots are possible.
4. Test likely candidates, starting with simple values such as \(\pm 1\), \(\pm 2\), or fractions suggested by the theorem.
5. When you find a root, divide to reduce the polynomial.
6. Factor the remaining polynomial if possible.
7. Check that your final roots make sense with Descartes' Rule and the degree of the polynomial.

11. Brief Summary

The Rational Root Theorem gives a complete list of all possible rational zeros of a polynomial by using factors of the constant term and the leading coefficient.

Descartes' Rule of Signs predicts how many positive and negative real zeros a polynomial may have by counting sign changes in \(f(x)\) and \(f(-x)\).

Used together, these tools make polynomial factorization much more systematic. Instead of guessing roots randomly, you can build a list, predict the signs of real roots, and test candidates efficiently.

Fundamental Theorem of Algebra is one of the most important ideas in polynomial math. It tells us how many roots a polynomial must have. This helps us factor polynomials, solve polynomial equations, and understand the shape of polynomial graphs.

In this lesson, you will learn what the theorem says, what counts as a root, why complex numbers matter, and how multiplicity affects the total number of roots.

Recall that a polynomial is an expression like

$$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,$$

where the exponents are whole numbers and the leading coefficient satisfies \(a_n\neq 0\). The highest exponent, \(n\), is called the degree of the polynomial.

A root or zero of a polynomial is a number \(r\) such that \(P(r)=0\).

The Fundamental Theorem of Algebra says:

$$\text{Every non-constant polynomial of degree } n \text{ has exactly } n \text{ complex roots, counting multiplicity.}$$

This statement has two very important parts:

• Every non-constant polynomial has at least one root.
• A polynomial of degree \(n\) has exactly \(n\) roots when complex roots and repeated roots are included.

This means a quadratic has 2 roots, a cubic has 3 roots, a 4th-degree polynomial has 4 roots, and so on, as long as we count correctly.

At first, this may seem different from what you have seen before. For example, the equation \(x^2+1=0\) appears to have no real solution, because no real number squared gives \(-1\). But in the complex number system, it has two roots:

$$x=i \quad \text{and} \quad x=-i.$$

So the theorem works only when we allow complex numbers.

Complex numbers are numbers of the form

$$a+bi,$$

where \(a\) and \(b\) are real numbers and \(i^2=-1\).

Because of the Fundamental Theorem of Algebra, every polynomial can be written as a product of linear factors over the complex numbers:

$$P(x)=a(x-r_1)(x-r_2)\cdots(x-r_n),$$

where \(a\) is the leading coefficient and the \(r_i\) values are the roots.

Sometimes some of the roots are the same. This is where multiplicity matters.

If a factor repeats, then its root repeats. For example, in

$$P(x)=(x-2)^3(x+1),$$

the root \(x=2\) appears 3 times, so it has multiplicity 3. The root \(x=-1\) appears once, so it has multiplicity 1.

The total number of roots, counting multiplicity, is

$$3+1=4,$$

which matches the degree of the polynomial.

This idea is extremely important. If a degree 5 polynomial seems to have only 3 roots, then some roots must be repeated, complex, or both.

Real roots vs. complex roots is another key idea.

• A polynomial may have all real roots.
• A polynomial may have some real roots and some complex roots.
• A polynomial may have no real roots at all, but it still has complex roots.

If a polynomial has real coefficients, then any non-real complex roots come in conjugate pairs.

That means if \(a+bi\) is a root, then \(a-bi\) must also be a root.

For example, if \(2+3i\) is a root of a polynomial with real coefficients, then \(2-3i\) must also be a root.

This helps us find missing roots.

How this connects to graphing

When you graph a polynomial on the real coordinate plane, you only see the real roots as x-intercepts. Complex roots do not appear as x-intercepts on the usual graph.

So a graph may show fewer x-intercepts than the degree of the polynomial. The “missing” roots are complex or repeated roots.

For example, the polynomial

$$f(x)=x^4+1$$

has degree 4, so it must have 4 complex roots. But its graph never crosses the x-axis, so it has no real roots. All 4 roots are complex.

Important facts to remember

• A polynomial of degree \(n\) has exactly \(n\) complex roots, counting multiplicity.
• Some of those roots may be real.
• Some may be repeated.
• Some may be non-real complex numbers.
• If the coefficients are real, non-real complex roots occur in conjugate pairs.

Now let’s work through examples.

Worked Example 1: A polynomial with all real roots

Find all roots of

$$P(x)=x^3-6x^2+11x-6.$$

We try factoring:

$$P(x)=(x-1)(x-2)(x-3).$$

So the roots are

$$x=1,\;2,\;3.$$

This is a degree 3 polynomial, and we found 3 roots. That matches the Fundamental Theorem of Algebra.

Worked Example 2: A polynomial with repeated roots

Find all roots of

$$P(x)=x^4-4x^3+6x^2-4x+1.$$

This polynomial factors as

$$P(x)=(x-1)^4.$$

So the only root value is \(x=1\), but it occurs 4 times.

The roots are

$$x=1,\;1,\;1,\;1.$$

We say \(x=1\) has multiplicity 4.

Even though there is only one distinct root, the polynomial still has exactly 4 roots counting multiplicity, which agrees with the theorem.

Worked Example 3: A polynomial with complex roots

Find all roots of

$$P(x)=x^2+4x+13.$$

Use the quadratic formula:

$$x=\frac{-4\pm\sqrt{4^2-4(1)(13)}}{2(1)}=\frac{-4\pm\sqrt{16-52}}{2}.$$

Simplify:

$$x=\frac{-4\pm\sqrt{-36}}{2}=\frac{-4\pm 6i}{2}=-2\pm 3i.$$

So the roots are

$$x=-2+3i \quad \text{and} \quad x=-2-3i.$$

This degree 2 polynomial has exactly 2 roots, but they are not real. Since the coefficients are real, the roots form a conjugate pair.

Worked Example 4: Using the theorem to find missing roots

A polynomial has degree 5, real coefficients, and roots \(2\), \(-1\), and \(3i\). Find the remaining roots.

Because the coefficients are real and \(3i\) is a non-real complex root, \(-3i\) must also be a root.

So far we have 4 roots:

$$2,\;-1,\;3i,\;-3i.$$

A degree 5 polynomial must have exactly 5 roots counting multiplicity, so there must be one more root.

If no other information is given, we can only say that there is one additional root. If the listed roots were all distinct and complete except for one, then the fifth root is simply some other root value.

However, if the question means these are the known roots and asks what root is guaranteed by conjugates, then the missing root determined by the theorem and conjugate rule is

$$-3i.$$

This example shows how the theorem helps us count roots correctly.

How to apply the Fundamental Theorem of Algebra

1. Find the degree of the polynomial.
2. Know that the total number of roots must equal the degree.
3. Include repeated roots according to multiplicity.
4. Include complex roots, even if they do not appear on the graph.
5. If the coefficients are real, remember that non-real complex roots come in conjugate pairs.

Common mistakes

• Counting only real roots: A polynomial may have fewer real roots than its degree, but it still has enough complex roots to make the total equal the degree.
• Ignoring multiplicity: If \((x-2)^3\) is a factor, then \(x=2\) counts as 3 roots.
• Forgetting conjugates: For real-coefficient polynomials, if \(4+i\) is a root, then \(4-i\) must also be a root.
• Thinking the graph shows every root: A real graph only shows real x-intercepts, not complex roots.

Quick check

• Degree 3 polynomial: total roots? 3
• Degree 6 polynomial: total roots? 6
• Polynomial factor \((x+5)^2\): root and multiplicity? Root \(-5\), multiplicity 2
• If \(1-2i\) is a root of a polynomial with real coefficients, what else is a root? \(1+2i\)

Brief Summary

The Fundamental Theorem of Algebra says that every non-constant polynomial of degree \(n\) has exactly \(n\) complex roots when counted with multiplicity. Some roots may be real, some may be repeated, and some may be complex. This theorem helps you solve polynomial equations, factor polynomials, and understand why the number of roots always matches the degree.

Root Multiplicity and Curve Behavior

When you graph a polynomial, the zeros of the polynomial are the x-values where the graph meets the x-axis. But not all zeros behave the same way.

Some zeros make the graph cross the x-axis, while others make the graph only touch the x-axis and turn around. This difference is explained by the multiplicity of the root.

Understanding multiplicity helps you sketch polynomial graphs more accurately and explain their behavior near each x-intercept.

1. What is a root with multiplicity?

If a polynomial has a factor of the form \((x-a)^n\), then \(x=a\) is a root with multiplicity \(n\).

For example:

• \((x-2)\) means \(x=2\) has multiplicity 1
• \((x+3)^2\) means \(x=-3\) has multiplicity 2
• \((x-1)^4\) means \(x=1\) has multiplicity 4

The multiplicity tells us how many times that root is repeated in the factorization.

2. How multiplicity affects the graph

The most important rule is this:

• If a root has odd multiplicity, the graph crosses the x-axis at that root.
• If a root has even multiplicity, the graph is tangent to the x-axis there, meaning it touches the axis and turns around.

This happens because of the sign of the polynomial near the root.

At an odd multiplicity root, the sign of the function changes from positive to negative or from negative to positive, so the graph must cross the x-axis.

At an even multiplicity root, the sign does not change, so the graph stays on the same side of the x-axis and only touches it.

3. Visual idea of different multiplicities

• Multiplicity 1: crosses straight through the x-axis
• Multiplicity 2: touches the x-axis and bounces back
• Multiplicity 3: crosses the x-axis, but more flatly than multiplicity 1
• Multiplicity 4: touches the x-axis and turns around, with an even flatter shape

So, multiplicity tells you both whether the graph crosses or touches, and how sharp or flat the graph looks near that root.

4. Why odd and even multiplicity behave differently

Consider the factor \((x-a)^n\).

If \(n\) is even, then values of \((x-a)^n\) are positive on both sides of \(x=a\), except exactly at the root where the value is 0. That means the graph does not switch sides of the x-axis.

If \(n\) is odd, then \((x-a)^n\) is negative on one side of \(x=a\) and positive on the other side. So the graph changes sign and crosses the x-axis.

5. Finding multiplicity from factored form

If the polynomial is already factored, you can read the roots and multiplicities directly.

Example:

$$f(x)=(x-4)^2(x+1)^3(x-2)$$

• Root \(x=4\), multiplicity 2
• Root \(x=-1\), multiplicity 3
• Root \(x=2\), multiplicity 1

Now use the rule:

• At \(x=4\), multiplicity 2 is even, so the graph touches and turns around.
• At \(x=-1\), multiplicity 3 is odd, so the graph crosses the x-axis.
• At \(x=2\), multiplicity 1 is odd, so the graph crosses the x-axis.

6. Worked Example 1

Determine how the graph behaves at each zero of

$$f(x)=(x-3)(x+2)^2$$

Step 1: Find the zeros.

• \(x=3\)
• \(x=-2\)

Step 2: Find the multiplicities.

• \((x-3)\) gives multiplicity 1
• \((x+2)^2\) gives multiplicity 2

Step 3: Describe the graph behavior.

• At \(x=3\), the multiplicity is odd, so the graph crosses the x-axis.
• At \(x=-2\), the multiplicity is even, so the graph touches the x-axis and turns around.

Conclusion: The graph is tangent to the x-axis at \(x=-2\) and crosses at \(x=3\).

7. Worked Example 2

Analyze the zeros of

$$g(x)=-(x+1)^3(x-4)^2$$

Step 1: Identify the zeros and multiplicities.

• \(x=-1\), multiplicity 3
• \(x=4\), multiplicity 2

Step 2: Determine the behavior at each root.

• Multiplicity 3 is odd, so at \(x=-1\) the graph crosses the x-axis.
• Multiplicity 2 is even, so at \(x=4\) the graph touches the x-axis and turns around.

Step 3: Notice the negative sign in front.

The negative sign affects whether the graph is reflected across the x-axis, but it does not change whether the graph crosses or touches at a root. That depends only on the multiplicity.

Conclusion: The graph crosses at \(x=-1\) and is tangent at \(x=4\).

8. Worked Example 3

For the polynomial

$$h(x)=(x-2)^4(x+3)$$

describe the behavior at each zero and explain which root looks flatter on the graph.

Step 1: Find the roots and multiplicities.

• \(x=2\), multiplicity 4
• \(x=-3\), multiplicity 1

Step 2: Use odd/even multiplicity.

• At \(x=2\), multiplicity 4 is even, so the graph touches the x-axis and turns around.
• At \(x=-3\), multiplicity 1 is odd, so the graph crosses the x-axis.

Step 3: Compare the shapes.

The root at \(x=2\) has higher multiplicity, so the graph will look flatter there than at \(x=-3\). A multiplicity 4 root usually creates a wide, gentle touch at the x-axis.

Conclusion: The graph crosses at \(x=-3\), touches at \(x=2\), and is flatter at \(x=2\).

9. Worked Example 4

Use root multiplicity to help sketch the x-axis behavior of

$$p(x)=(x+2)^2(x-1)^3(x-5)$$

Step 1: List the roots.

• \(x=-2\), multiplicity 2
• \(x=1\), multiplicity 3
• \(x=5\), multiplicity 1

Step 2: Decide how the graph behaves at each one.

• At \(x=-2\), even multiplicity means the graph touches and turns around.
• At \(x=1\), odd multiplicity means the graph crosses the x-axis.
• At \(x=5\), odd multiplicity means the graph crosses the x-axis.

Step 3: Compare the crossings.

The graph crosses at both \(x=1\) and \(x=5\), but it will usually look flatter at \(x=1\) because multiplicity 3 is greater than multiplicity 1.

Conclusion: When sketching, show a bounce at \(x=-2\), a flatter crossing at \(x=1\), and a more direct crossing at \(x=5\).

10. Important note when sketching full polynomial graphs

Root multiplicity tells you what happens at the x-intercepts, but to sketch the entire graph you may also need:

• the degree of the polynomial
• the leading coefficient
• the end behavior
• possibly the y-intercept

Still, multiplicity is one of the most important tools because it tells you exactly how the graph behaves where it meets the x-axis.

11. Common mistakes to avoid

• Confusing the root with the factor: If the factor is \((x-4)^2\), the root is \(x=4\), not \(x=-4\).
• Ignoring multiplicity: Two roots may both be x-intercepts, but one may cross while the other only touches.
• Thinking all odd multiplicities look the same: They all cross, but higher odd multiplicities look flatter.
• Thinking the coefficient changes crossing/touching behavior: The sign or leading coefficient may flip the graph, but even/odd multiplicity still decides whether it crosses or touches.

12. Quick rule chart

• If the factor is \((x-a)^n\), then \(x=a\) is a root of multiplicity \(n\).
• If \(n\) is odd, the graph crosses the x-axis.
• If \(n\) is even, the graph touches the x-axis and turns around.
• Larger multiplicity means the graph is usually flatter near that root.

13. Summary

Root multiplicity tells you how a polynomial graph behaves at each zero. If a root has odd multiplicity, the graph crosses the x-axis. If a root has even multiplicity, the graph touches the x-axis and turns around.

Higher multiplicities make the graph appear flatter near the root. When a polynomial is written in factored form, you can find each root, identify its multiplicity, and use that information to sketch the graph more accurately.

Vertical Asymptotes and Infinite Limits

When you graph a rational function, sometimes the graph shoots upward or downward without bound near a certain value of \(x\). This behavior is closely connected to places where the function is undefined. In this lesson, you will learn how to find vertical asymptotes, how to describe what the function does near them using infinite limits, and how this helps you sketch the graph correctly.

A rational function is a function of the form

where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)\neq 0\).

Since division by zero is impossible, the values of \(x\) that make the denominator zero are not in the domain. Some of these excluded values create vertical asymptotes.

Definition: A vertical asymptote is a vertical line \(x=a\) such that the function values become extremely large positive or negative as \(x\) gets very close to \(a\).

We write this using limits:

Here:

• \(x\to a^-\) means \(x\) approaches \(a\) from the left.
• \(x\to a^+\) means \(x\) approaches \(a\) from the right.
• \(\infty\) and \(-\infty\) mean the function grows without bound upward or downward.

Important idea: Not every denominator zero gives a vertical asymptote. Sometimes a factor cancels, and the graph has a hole instead.

So, to find vertical asymptotes, follow this process.

1. Factor the numerator and denominator completely.
2. Identify any common factors.
3. Cancel common factors carefully.
4. Set the remaining denominator equal to zero.
5. The solutions are the vertical asymptotes.

If a factor cancels completely, that value of \(x\) is usually a hole, not a vertical asymptote.

How to tell whether the graph goes up or down near a vertical asymptote

After finding the asymptote \(x=a\), test values just to the left and right of \(a\), or reason from the signs of the factors.

For example, if near \(x=a\), the numerator is positive and the denominator is a very small positive number, then the quotient is a very large positive number. If the denominator is a very small negative number, then the quotient is a very large negative number.

This sign analysis lets you determine the one-sided infinite limits.

Example 1: A basic vertical asymptote

Find the vertical asymptote(s) and infinite limits of

Step 1: Find where the denominator is zero.

So \(x=3\) is a possible vertical asymptote. There are no common factors to cancel, so it is a vertical asymptote.

Step 2: Check behavior on each side of \(x=3\).

• If \(x\to 3^-\), then \(x-3\) is a very small negative number.
• So \(\frac{1}{x-3}\) becomes a very large negative number.

• If \(x\to 3^+\), then \(x-3\) is a very small positive number.
• So \(\frac{1}{x-3}\) becomes a very large positive number.

Conclusion: The function has a vertical asymptote at \(x=3\). The graph falls to \(-\infty\) on the left side and rises to \(\infty\) on the right side.

Example 2: A rational function with factoring

Find the vertical asymptote(s) and infinite limits of

Step 1: Factor the denominator.

So

Step 2: Cancel common factors.

The factor \((x+2)\) cancels, so \(x=-2\) is not a vertical asymptote. It is a hole.

The remaining denominator is zero when

So the only vertical asymptote is \(x=2\).

Step 3: Find the one-sided limits near \(x=2\).

• As \(x\to 2^-\), \(x-2\) is a small negative number, so \(\frac{1}{x-2}\to -\infty\).
• As \(x\to 2^+\), \(x-2\) is a small positive number, so \(\frac{1}{x-2}\to \infty\).

Conclusion: The graph has a hole at \(x=-2\) and a vertical asymptote at \(x=2\).

Example 3: Determining the direction on both sides

Find the vertical asymptote(s) and infinite limits of

Step 1: Find denominator zeros.

There are no common factors to cancel, so \(x=4\) is a vertical asymptote.

Step 2: Analyze the sign near \(x=4\).

The denominator \((x-4)^2\) is always positive except at \(x=4\), where it is zero. Near \(x=4\), it is a very small positive number.

The numerator near \(x=4\) is \(x+1\), which is about \(5\), so it is positive.

Positive divided by a very small positive number gives a very large positive number on both sides.

Conclusion: The graph rises to \(\infty\) on both sides of the vertical asymptote \(x=4\).

Example 4: Two vertical asymptotes

Find the vertical asymptote(s) and infinite limits of

Step 1: Factor the denominator.

So

There are no common factors to cancel, so the vertical asymptotes are

Step 2: Analyze near \(x=1\).

Near \(x=1\):

• \(2x\) is positive.
• \(x+1\) is positive.
• The sign depends mainly on \(x-1\).

If \(x\to 1^-\), then \(x-1\) is negative, so the denominator is negative. A positive numerator divided by a very small negative denominator gives \(-\infty\).

If \(x\to 1^+\), then \(x-1\) is positive, so the denominator is positive. A positive numerator divided by a very small positive denominator gives \(\infty\).

Step 3: Analyze near \(x=-1\).

Near \(x=-1\):

• \(2x\) is negative.
• \(x-1\) is negative.
• The sign depends mainly on \(x+1\).

If \(x\to -1^-\), then \(x+1\) is negative. The denominator is negative times negative, which is positive. A negative numerator divided by a very small positive denominator gives \(-\infty\).

If \(x\to -1^+\), then \(x+1\) is positive. The denominator is negative times positive, which is negative. A negative numerator divided by a very small negative denominator gives \(\infty\).

Conclusion: There are vertical asymptotes at \(x=-1\) and \(x=1\), and the graph changes direction differently at each side depending on the signs.

Key ideas to remember

• Vertical asymptotes occur at values of \(x\) that make the remaining denominator zero after simplification.
• If a factor cancels, that value usually gives a hole, not a vertical asymptote.
• Use one-sided limits to describe what happens on the left and right of the asymptote.
• If the function approaches \(\infty\) or \(-\infty\), the limit is an infinite limit.
• The sign of the numerator and denominator near the asymptote tells whether the graph goes up or down.

Common mistakes

• Mistake 1: Forgetting to factor completely before deciding on asymptotes.
• Mistake 2: Treating a canceled factor as a vertical asymptote instead of a hole.
• Mistake 3: Writing only \(\lim_{x\to a}f(x)\) when the left and right sides behave differently. In that case, use one-sided limits.
• Mistake 4: Not checking signs carefully near the asymptote.

How this helps with graphing

When sketching a rational function, vertical asymptotes tell you where the graph is broken into separate pieces. The one-sided infinite limits tell you whether each piece rises or falls near the asymptote.

So when you graph:

1. Find the domain restrictions.
2. Factor and simplify.
3. Identify holes and vertical asymptotes.
4. Use one-sided limits to determine the behavior near each asymptote.
5. Add intercepts and other features to finish the sketch.

Brief Summary

A vertical asymptote is a line \(x=a\) where a rational function grows without bound because the denominator approaches zero. To find vertical asymptotes, factor first, cancel any common factors, and then set the remaining denominator equal to zero. Infinite limits describe whether the function goes to \(\infty\) or \(-\infty\) from the left or right side. This information is essential for sketching rational graphs accurately.

Horizontal and Oblique Asymptotes are important tools for understanding the end behavior of rational functions. A rational function is a function of the form

$$f(x)=\frac{P(x)}{Q(x)}$$

where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)\neq 0\).

Asymptotes help us describe what happens to the graph when \(x\) becomes very large positive or very large negative. In this lesson, we will focus on two types:

• Horizontal asymptotes
• Oblique asymptotes (also called slant asymptotes)

These asymptotes tell us what the graph approaches far to the left and far to the right.

1. What is a horizontal asymptote?

A horizontal asymptote is a horizontal line \(y=L\) that the graph gets closer to as \(x\to \infty\) or \(x\to -\infty\).

In limit form, if

$$\lim_{x\to \infty} f(x)=L \quad \text{or} \quad \lim_{x\to -\infty} f(x)=L,$$

then \(y=L\) is a horizontal asymptote.

For rational functions, horizontal asymptotes can often be found just by comparing the degrees of the numerator and denominator.

2. Compare the degrees of numerator and denominator

Suppose

$$f(x)=\frac{P(x)}{Q(x)}$$

and let:

• degree of numerator = \(n\)
• degree of denominator = \(m\)

Then the end behavior depends on the relationship between \(n\) and \(m\).

Case A: Degree of numerator is less than degree of denominator

If \(n<m\), then

$$y=0$$

is the horizontal asymptote.

This happens because the denominator grows faster than the numerator, so the fraction gets closer and closer to 0.

Case B: Degree of numerator equals degree of denominator

If \(n=m\), then the horizontal asymptote is

$$y=\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}.$$

The leading coefficients are the coefficients of the highest power terms.

Case C: Degree of numerator is greater than degree of denominator

If \(n>m\), then there is no horizontal asymptote.

However, if the degree of the numerator is exactly one more than the degree of the denominator, then the function may have an oblique asymptote.

3. What is an oblique asymptote?

An oblique asymptote is a slanted line of the form

$$y=mx+b$$

that the graph approaches as \(x\to \infty\) or \(x\to -\infty\).

A rational function has an oblique asymptote when:

$$\text{degree of numerator} = \text{degree of denominator} + 1.$$

To find the oblique asymptote, divide the numerator by the denominator using polynomial long division or synthetic division if appropriate.

The quotient without the remainder gives the asymptote.

If

$$f(x)=\frac{P(x)}{Q(x)}=\text{quotient}+\frac{\text{remainder}}{Q(x)},$$

then as \(x\) becomes very large, the fraction

$$\frac{\text{remainder}}{Q(x)}$$

gets very small, so the graph approaches the quotient line.

4. Why degree comparison works

For large values of \(x\), the highest power terms matter the most. Lower-degree terms become much less important.

For example, in

$$\frac{3x^2+5x-1}{x^2-4},$$

the terms \(3x^2\) and \(x^2\) control the end behavior. Since

$$\frac{3x^2}{x^2}=3,$$

the graph approaches \(y=3\).

This is why the ratio of leading coefficients gives the horizontal asymptote when the degrees are equal.

5. Worked Example 1: Numerator degree less than denominator degree

Find the horizontal asymptote of

$$f(x)=\frac{2x+1}{x^2+3}.$$

Step 1: Compare degrees.

• Degree of numerator = 1
• Degree of denominator = 2

Since \(1<2\), the numerator degree is less than the denominator degree.

Step 2: Apply the rule.

So the horizontal asymptote is

$$y=0.$$

Answer: The function has horizontal asymptote \(y=0\).

6. Worked Example 2: Degrees are equal

Find the horizontal asymptote of

$$f(x)=\frac{4x^3-2x+7}{2x^3+5x^2-1}.$$

Step 1: Compare degrees.

• Degree of numerator = 3
• Degree of denominator = 3

The degrees are equal.

Step 2: Find the leading coefficients.

• Leading coefficient of numerator = 4
• Leading coefficient of denominator = 2

Step 3: Form the ratio.

$$y=\frac{4}{2}=2$$

Answer: The horizontal asymptote is

$$y=2.$$

7. Worked Example 3: Finding an oblique asymptote

Find the oblique asymptote of

$$f(x)=\frac{x^2+3x+2}{x+1}.$$

Step 1: Compare degrees.

• Degree of numerator = 2
• Degree of denominator = 1

Since the numerator degree is exactly one more than the denominator degree, the function has an oblique asymptote.

Step 2: Divide the polynomials.

Divide \(x^2+3x+2\) by \(x+1\):

$$\frac{x^2+3x+2}{x+1}=x+2+\frac{0}{x+1}$$

because

$$x^2+3x+2=(x+1)(x+2).$$

Step 3: Use the quotient.

The quotient is \(x+2\), so the oblique asymptote is

$$y=x+2.$$

Answer: The oblique asymptote is \(y=x+2\).

In this example, the graph actually lies on the line except where the function is undefined, but the line is still the oblique asymptote.

8. Worked Example 4: Oblique asymptote with a remainder

Find the oblique asymptote of

$$f(x)=\frac{2x^2+5x+1}{x+2}.$$

Step 1: Compare degrees.

• Degree of numerator = 2
• Degree of denominator = 1

Again, the numerator degree is one more than the denominator degree, so there is an oblique asymptote.

Step 2: Use polynomial division.

Divide \(2x^2+5x+1\) by \(x+2\).

$$2x^2+5x+1 \div (x+2)$$

The division gives

$$\frac{2x^2+5x+1}{x+2}=2x+1-\frac{1}{x+2}.$$

Step 3: Identify the asymptote.

As \(x\to \infty\) or \(x\to -\infty\), the term

$$-\frac{1}{x+2}$$

approaches 0.

So the graph approaches

$$y=2x+1.$$

Answer: The oblique asymptote is

$$y=2x+1.$$

9. Important notes and common mistakes

• Do not find horizontal asymptotes by plugging in a large number only once. Use degree comparison or limits of end behavior.
• A function cannot have both a horizontal and an oblique asymptote. The degree relationship gives one situation or the other.
• Oblique asymptotes only happen when the numerator degree is exactly one more than the denominator degree.
• Horizontal asymptotes describe end behavior, not where the graph must stop. A graph can cross a horizontal asymptote.

For example, a rational function may cross its horizontal asymptote at some value of \(x\), but still approach that line as \(x\to \infty\) or \(x\to -\infty\).

10. Quick rule chart

• If \(\deg(\text{numerator}) < \deg(\text{denominator})\), then horizontal asymptote: \(y=0\)
• If \(\deg(\text{numerator}) = \deg(\text{denominator})\), then horizontal asymptote: ratio of leading coefficients
• If \(\deg(\text{numerator}) = \deg(\text{denominator})+1\), then oblique asymptote: use polynomial division
• If \(\deg(\text{numerator}) > \deg(\text{denominator})+1\), then there is no horizontal asymptote and no oblique asymptote of the form \(y=mx+b\)

11. How this helps with graphing

When sketching a rational graph, asymptotes help you understand the overall shape.

Horizontal asymptotes show the value the graph levels off toward far left and far right.

Oblique asymptotes show the slanted line the graph follows at the ends.

These are especially useful when you combine them with:

• x-intercepts
• y-intercepts
• vertical asymptotes
• holes, if any factors cancel

12. Brief summary

To find horizontal and oblique asymptotes of a rational function, compare the degrees of the numerator and denominator.

If the numerator degree is smaller, the horizontal asymptote is \(y=0\). If the degrees are equal, the horizontal asymptote is the ratio of leading coefficients.

If the numerator degree is exactly one more than the denominator degree, use polynomial division to find the oblique asymptote.

These asymptotes describe the end behavior of the graph and are a key part of sketching rational functions correctly.

Removable discontinuities are a special type of break in the graph of a rational function. They happen when the numerator and denominator share a common factor, and that factor can be canceled.

When this happens, the graph looks like it follows a smooth curve, but one point is missing. That missing point is called a hole.

This idea is important when graphing rational functions because not every discontinuity creates a vertical asymptote. Some discontinuities are removable, meaning the function could be “fixed” by defining the missing point correctly.

1. What is a rational function?

A rational function is a function that can be written as

$$f(x)=\frac{P(x)}{Q(x)}$$

where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)\neq 0\).

A function is discontinuous at values of \(x\) that make the denominator equal to 0. But there are different kinds of discontinuities.

• If a factor in the denominator does not cancel, the graph usually has a vertical asymptote.
• If a factor in the denominator does cancel with the numerator, the graph has a removable discontinuity, or hole.

2. Why is it called “removable”?

Suppose a function has the form

$$f(x)=\frac{(x-2)(x+3)}{x-2}$$

For every value of \(x\) except \(x=2\), this simplifies to

$$f(x)=x+3$$

But the original function is still undefined at \(x=2\), because the denominator becomes 0 there.

So the graph is the line \(y=x+3\), except the point where \(x=2\) is missing. Since the simplified expression would give \(y=5\), the hole is at

$$ (2,5) $$

This discontinuity is called removable because if we defined the function value at \(x=2\) to be 5, the graph would become continuous.

3. How to identify a removable discontinuity

To find a removable discontinuity in a rational function, follow these steps:

1. Factor the numerator and denominator completely.
2. Look for common factors.
3. Cancel any common factors.
4. The value of \(x\) that makes the canceled factor zero is the x-value of the hole.
5. Substitute that \(x\)-value into the simplified function to find the y-value of the hole.

Important: The hole comes from the original function being undefined, even though the simplified function exists there.

4. Worked Example 1: A basic hole

Find any removable discontinuity of

$$f(x)=\frac{x^2-9}{x-3}$$

Step 1: Factor the numerator.

$$x^2-9=(x-3)(x+3)$$

So

$$f(x)=\frac{(x-3)(x+3)}{x-3}$$

Step 2: Cancel the common factor.

$$f(x)=x+3 \quad \text{for } x\neq 3$$

Step 3: Find the x-value of the hole.

The canceled factor is \(x-3\), so the hole occurs at

$$x=3$$

Step 4: Find the y-value of the hole.

Use the simplified function:

$$y=3+3=6$$

Answer: The graph has a removable discontinuity, or hole, at

$$ (3,6) $$

5. Worked Example 2: Hole or vertical asymptote?

Determine the discontinuities of

$$f(x)=\frac{x^2-4}{x^2-x-6}$$

Step 1: Factor completely.

$$x^2-4=(x-2)(x+2)$$

$$x^2-x-6=(x-3)(x+2)$$

So

$$f(x)=\frac{(x-2)(x+2)}{(x-3)(x+2)}$$

Step 2: Cancel common factors.

$$f(x)=\frac{x-2}{x-3} \quad \text{for } x\neq -2,3$$

Step 3: Identify the discontinuities.

• The canceled factor \((x+2)\) gives a hole at \(x=-2\).
• The remaining denominator \((x-3)\) gives a vertical asymptote at \(x=3\).

Step 4: Find the coordinates of the hole.

Substitute \(x=-2\) into the simplified function:

$$y=\frac{-2-2}{-2-3}=\frac{-4}{-5}=\frac{4}{5}$$

Answer:

• Removable discontinuity at $$(-2,\tfrac{4}{5})$$
• Vertical asymptote at $$x=3$$

This example shows that one rational function can have both a hole and a vertical asymptote.

6. Worked Example 3: A hole after more factoring

Find the removable discontinuity of

$$f(x)=\frac{x^2+5x+6}{x^2+4x+3}$$

Step 1: Factor both polynomials.

$$x^2+5x+6=(x+2)(x+3)$$

$$x^2+4x+3=(x+1)(x+3)$$

So

$$f(x)=\frac{(x+2)(x+3)}{(x+1)(x+3)}$$

Step 2: Cancel the common factor.

$$f(x)=\frac{x+2}{x+1} \quad \text{for } x\neq -3,-1$$

Step 3: Identify the hole.

The canceled factor is \((x+3)\), so the hole occurs at

$$x=-3$$

Step 4: Find the y-value.

$$y=\frac{-3+2}{-3+1}=\frac{-1}{-2}=\frac{1}{2}$$

Answer: The removable discontinuity is at

$$(-3,\tfrac{1}{2})$$

The function also has a vertical asymptote at \(x=-1\), because that factor remains in the denominator.

7. How holes appear on a graph

When a rational function has a removable discontinuity, the graph follows the simplified function everywhere except at the missing x-value.

To sketch the graph:

1. Simplify the function by canceling common factors.
2. Graph the simplified expression.
3. Find the x-value from the canceled factor.
4. Find the corresponding y-value from the simplified function.
5. Draw an open circle at that point to show the hole.

An open circle means the point is not included in the graph.

8. Common mistakes to avoid

• Not factoring completely. If you do not factor fully, you may miss a common factor and fail to find a hole.
• Assuming every denominator zero gives a vertical asymptote. First check whether the factor cancels.
• Using the original function to find the y-value of the hole. The original function is undefined there. Use the simplified function instead.
• Forgetting the point is missing. Even though the simplified expression gives a y-value, the graph must show an open circle at the hole.

9. Quick comparison: Hole vs. vertical asymptote

• Hole: caused by a factor that appears in both numerator and denominator and cancels.
• Vertical asymptote: caused by a factor that stays in the denominator after simplification.

For example, in

$$\frac{(x-1)(x+4)}{(x-1)(x-2)}$$

the factor \((x-1)\) cancels, so there is a hole at \(x=1\). The factor \((x-2)\) remains in the denominator, so there is a vertical asymptote at \(x=2\).

10. Practice thinking

Ask yourself these questions when you see a rational function:

• What values make the denominator zero?
• Do any of those factors cancel with the numerator?
• If a factor cancels, where is the hole?
• If a factor does not cancel, is there a vertical asymptote?

This process helps you understand the graph correctly.

11. Brief summary

A removable discontinuity happens when a rational function has a common factor in the numerator and denominator. After canceling that factor, the graph matches the simplified function, but the x-value from the canceled factor becomes a missing point, or hole.

To find the hole, factor completely, cancel common factors, use the canceled factor to get the x-value, and plug that value into the simplified function to get the y-value. On the graph, show the hole with an open circle.

Sketching Rational Functions means putting together several graph features so you can draw a reasonable and accurate graph by hand.

A rational function is a function that can be written as

$$f(x)=\frac{P(x)}{Q(x)}$$

where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)\neq 0\).

To sketch a rational function well, you need to identify its domain, x-intercepts, y-intercept, vertical asymptotes, horizontal or slant asymptotes, any holes, and the sign of the function on different intervals.

This lesson will show you a clear step-by-step method and then apply it in worked examples.

1. Start by factoring completely

Factoring is the key first step because it helps you see zeros, undefined values, and possible simplifications.

For example, if

$$f(x)=\frac{(x-2)(x+1)}{(x-2)(x-3)}$$

then the factor \((x-2)\) appears in both numerator and denominator. That tells you there may be a hole at \(x=2\), not a vertical asymptote.

2. Find the domain

The domain of a rational function is all real numbers except values that make the denominator zero.

So if the denominator is zero at \(x=a\), then \(x=a\) is excluded from the domain.

Important: even if a factor cancels, the original denominator was still zero there, so that value is still not in the domain.

3. Identify holes

A hole occurs when a factor cancels from the numerator and denominator.

After canceling, substitute the excluded \(x\)-value into the simplified function to find the coordinates of the hole.

Example idea:

$$f(x)=\frac{(x-2)(x+1)}{(x-2)(x-3)}=\frac{x+1}{x-3},\quad x\neq 2,3$$

Since \(x=2\) was canceled, there is a hole at

$$\left(2,\frac{2+1}{2-3}\right)=(2,-3)$$

4. Find x-intercepts

The x-intercepts are where the graph crosses the \(x\)-axis, so they happen when \(f(x)=0\).

For a rational function, this means the numerator must be zero after removing any canceled factors.

Why? Because a canceled factor gives a hole, not an intercept.

So:

• Set the simplified numerator equal to 0.
• Solve for \(x\).
• Make sure the value is still in the domain.

5. Find the y-intercept

The y-intercept is found by substituting \(x=0\), as long as \(0\) is in the domain.

So compute \(f(0)\). The point is \((0,f(0))\).

6. Find vertical asymptotes

A vertical asymptote occurs where the denominator is zero after simplification.

That means:

• Factor first.
• Cancel common factors.
• Set the remaining denominator equal to 0.

If \(x=a\) makes the simplified denominator zero, then \(x=a\) is a vertical asymptote.

Near a vertical asymptote, the function values grow very large positively or negatively.

7. Find horizontal or slant asymptotes

These describe the graph's end behavior as \(x\to\infty\) or \(x\to -\infty\).

Compare the degrees of the numerator and denominator.

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y=0\).
• If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.
• If the degree of the numerator is one more than the degree of the denominator, there is a slant asymptote. Use polynomial division.

For Grade 12, these are the most common cases when sketching.

Example:

$$f(x)=\frac{2x^2+1}{x^2-3x+4}$$

The degrees are equal, so the horizontal asymptote is

$$y=\frac{2}{1}=2$$

8. Use sign chart analysis

A sign chart tells you where the function is positive and where it is negative.

This helps you know whether each branch of the graph lies above or below the \(x\)-axis.

To make a sign chart:

1. Find all critical \(x\)-values from the zeros of the numerator and denominator.
2. These values divide the number line into intervals.
3. Choose one test value from each interval.
4. Substitute into the factored form to determine the sign.

You do not need the exact function value each time. You only need to know whether each factor is positive or negative.

9. Check behavior near vertical asymptotes

It is not enough to say that \(x=a\) is a vertical asymptote. You should also know whether the graph goes to \(+\infty\) or \(-\infty\) on each side.

Use values just to the left and right of the asymptote, or use factor signs.

For example, if

$$f(x)=\frac{1}{x-2}$$

then:

• as \(x\to 2^-\), \(x-2\) is a very small negative number, so \(f(x)\to -\infty\)
• as \(x\to 2^+\), \(x-2\) is a very small positive number, so \(f(x)\to +\infty\)

10. Put everything together

Once you know the intercepts, holes, asymptotes, and signs, you can sketch the graph branch by branch.

A good sketch should show:

• all intercepts clearly
• vertical asymptotes as dashed vertical lines
• horizontal or slant asymptotes as dashed lines
• holes as open circles
• whether each branch is above or below the \(x\)-axis
• how the graph behaves near each asymptote

Step-by-step strategy for sketching any rational function

1. Factor numerator and denominator completely.
2. State the domain.
3. Cancel common factors and identify any holes.
4. Find the \(x\)-intercepts from the simplified numerator.
5. Find the \(y\)-intercept by substituting \(x=0\).
6. Find vertical asymptotes from the simplified denominator.
7. Find the horizontal or slant asymptote from the degrees.
8. Make a sign chart using zeros and vertical asymptotes.
9. Check one-sided behavior near vertical asymptotes.
10. Sketch the graph.

Worked Example 1: A basic rational function

Sketch

$$f(x)=\frac{x+1}{x-2}$$

Step 1: Factor

This function is already factored.

Step 2: Domain

The denominator is zero when \(x=2\), so

$$\text{Domain: }x\neq 2$$

Step 3: Holes

There are no common factors, so there are no holes.

Step 4: x-intercept

Set the numerator equal to zero:

$$x+1=0\Rightarrow x=-1$$

So the \(x\)-intercept is \((-1,0)\).

Step 5: y-intercept

$$f(0)=\frac{0+1}{0-2}= -\frac12$$

So the \(y\)-intercept is \((0,-\frac12)\).

Step 6: Vertical asymptote

Set denominator equal to zero:

$$x-2=0\Rightarrow x=2$$

So the vertical asymptote is

$$x=2$$

Step 7: Horizontal asymptote

The numerator and denominator both have degree 1. The ratio of leading coefficients is \(1/1=1\).

So the horizontal asymptote is

$$y=1$$

Step 8: Sign chart

Critical values are \(x=-1\) and \(x=2\).

• For \(x<-1\), choose \(x=-2\): \(\frac{-1}{-4}>0\), so positive.
• For \(-1<x<2\), choose \(x=0\): \(\frac{1}{-2}<0\), so negative.
• For \(x>2\), choose \(x=3\): \(\frac{4}{1}>0\), so positive.

Step 9: Behavior near the asymptote

• As \(x\to 2^-\), denominator is a small negative number, so \(f(x)\to -\infty\).
• As \(x\to 2^+\), denominator is a small positive number, so \(f(x)\to +\infty\).

Sketch description

The left branch crosses the \(x\)-axis at \((-1,0)\), passes through \((0,-\frac12)\), then falls to \(-\infty\) as it approaches \(x=2\) from the left.

The right branch is above the \(x\)-axis, rises from \(+\infty\) just right of \(x=2\), and approaches \(y=1\) as \(x\to\infty\).

Worked Example 2: Rational function with a hole

Sketch

$$f(x)=\frac{(x-1)(x+2)}{(x-1)(x-3)}$$

Step 1: Factor

It is already factored.

Step 2: Domain

The original denominator is zero at \(x=1\) and \(x=3\), so

$$\text{Domain: }x\neq 1,3$$

Step 3: Simplify and identify hole

Cancel the common factor \((x-1)\):

$$f(x)=\frac{x+2}{x-3},\quad x\neq 1,3$$

Because \(x=1\) was canceled, there is a hole there. Find its \(y\)-value using the simplified function:

$$y=\frac{1+2}{1-3}=\frac{3}{-2}=-\frac32$$

So the hole is at

$$\left(1,-\frac32\right)$$

Step 4: x-intercept

Set the simplified numerator equal to zero:

$$x+2=0\Rightarrow x=-2$$

So the \(x\)-intercept is \((-2,0)\).

Step 5: y-intercept

$$f(0)=\frac{(0-1)(0+2)}{(0-1)(0-3)}=\frac{-2}{3}$$

So the \(y\)-intercept is \((0,-\frac23)\).

Step 6: Vertical asymptote

After simplification, the denominator is zero at \(x=3\). So the vertical asymptote is

$$x=3$$

Step 7: Horizontal asymptote

The simplified function has equal degrees in numerator and denominator. Ratio of leading coefficients is \(1\).

So the horizontal asymptote is

$$y=1$$

Step 8: Sign chart

Use the simplified form \(\frac{x+2}{x-3}\). Critical values are \(-2\), \(1\), and \(3\). The hole at \(x=1\) still matters for the sketch because the graph is missing that point.

• For \(x<-2\), choose \(x=-3\): \(\frac{-1}{-6}>0\), positive.
• For \(-2<x<1\), choose \(x=0\): \(\frac{2}{-3}<0\), negative.
• For \(1<x<3\), choose \(x=2\): \(\frac{4}{-1}<0\), negative.
• For \(x>3\), choose \(x=4\): \(\frac{6}{1}>0\), positive.

Step 9: Behavior near \(x=3\)

• As \(x\to 3^-\), denominator is a small negative number, so \(f(x)\to -\infty\).
• As \(x\to 3^+\), denominator is a small positive number, so \(f(x)\to +\infty\).

Sketch description

The graph looks like \(\frac{x+2}{x-3}\), except it has an open circle at \(\left(1,-\frac32\right)\).

It crosses the \(x\)-axis at \((-2,0)\), has vertical asymptote \(x=3\), and approaches \(y=1\) for large positive and negative \(x\).

Worked Example 3: More than one vertical asymptote

Sketch

$$f(x)=\frac{x^2-1}{x^2-x-6}$$

Step 1: Factor

$$x^2-1=(x-1)(x+1)$$

$$x^2-x-6=(x-3)(x+2)$$

So

$$f(x)=\frac{(x-1)(x+1)}{(x-3)(x+2)}$$

Step 2: Domain

The denominator is zero at \(x=3\) and \(x=-2\), so

$$\text{Domain: }x\neq 3,-2$$

Step 3: Holes

There are no common factors, so there are no holes.

Step 4: x-intercepts

Set numerator equal to zero:

$$x-1=0\Rightarrow x=1$$

$$x+1=0\Rightarrow x=-1$$

So the \(x\)-intercepts are

$$(-1,0)\text{ and }(1,0)$$

Step 5: y-intercept

$$f(0)=\frac{-1}{-6}=\frac16$$

So the \(y\)-intercept is \((0,\frac16)\).

Step 6: Vertical asymptotes

The denominator is zero at \(x=-2\) and \(x=3\), so the vertical asymptotes are

$$x=-2\quad \text{and}\quad x=3$$

Step 7: Horizontal asymptote

The degrees are equal, and the leading coefficients are both 1. So the horizontal asymptote is

$$y=1$$

Step 8: Sign chart

Critical values are \(-2,-1,1,3\).

• For \(x<-2\), choose \(-3\): positive.
• For \(-2<x<-1\), choose \(-1.5\): negative.
• For \(-1<x<1\), choose \(0\): positive.
• For \(1<x<3\), choose \(2\): negative.
• For \(x>3\), choose \(4\): positive.

So the graph alternates above and below the \(x\)-axis across those intervals.

Step 9: One-sided behavior

Near \(x=-2\):

• as \(x\to -2^-\), \(f(x)\to +\infty\)
• as \(x\to -2^+\), \(f(x)\to -\infty\)

Near \(x=3\):

• as \(x\to 3^-\), \(f(x)\to -\infty\)
• as \(x\to 3^+\), \(f(x)\to +\infty\)

Sketch description

There are three branches: one left of \(-2\), one between \(-2\) and \(3\), and one right of \(3\).

The middle branch crosses the \(x\)-axis at \((-1,0)\) and \((1,0)\), passes through \((0,\frac16)\), and heads downward to \(-\infty\) as it approaches \(x=3\) from the left.

Worked Example 4: A slant asymptote

Sketch

$$f(x)=\frac{x^2+1}{x-1}$$

Step 1: Factor

The numerator does not factor nicely over the real numbers, so we leave it as it is.

Step 2: Domain

The denominator is zero at \(x=1\), so

$$\text{Domain: }x\neq 1$$

Step 3: Holes

There are no common factors, so there are no holes.

Step 4: x-intercepts

Set numerator equal to zero:

$$x^2+1=0$$

This has no real solutions, so there are no x-intercepts.

Step 5: y-intercept

$$f(0)=\frac{1}{-1}=-1$$

So the \(y\)-intercept is \((0,-1)\).

Step 6: Vertical asymptote

$$x-1=0\Rightarrow x=1$$

So the vertical asymptote is

$$x=1$$

Step 7: Slant asymptote

The degree of the numerator is one more than the degree of the denominator, so we divide:

$$\frac{x^2+1}{x-1}=x+1+\frac{2}{x-1}$$

As \(x\to\pm\infty\), the fraction \(\frac{2}{x-1}\to 0\), so the slant asymptote is

$$y=x+1$$

Step 8: Sign chart

Since \(x^2+1>0\) for all real \(x\), the sign depends only on \(x-1\).

• For \(x<1\), denominator is negative, so the function is negative.
• For \(x>1\), denominator is positive, so the function is positive.

Step 9: Behavior near \(x=1\)

• as \(x\to 1^-\), \(f(x)\to -\infty\)
• as \(x\to 1^+\), \(f(x)\to +\infty\)

Sketch description

The graph has one branch below the \(x\)-axis for \(x<1\), passing through \((0,-1)\), and one branch above the \(x\)-axis for \(x>1\).

Both branches get closer to the line \(y=x+1\) as \(x\) becomes very large positive or negative.

Common mistakes to avoid

• Forgetting to factor first. If you do not factor, you may miss holes or asymptotes.
• Using canceled factors as x-intercepts. A canceled factor gives a hole, not an intercept.
• Using the original denominator for vertical asymptotes after canceling. Only remaining denominator factors create vertical asymptotes.
• Ignoring sign analysis. Without it, your graph may be on the wrong side of the \(x\)-axis.
• Forgetting one-sided behavior near vertical asymptotes. The branch direction matters.
• Drawing the graph through an asymptote. A graph approaches an asymptote but does not cross a vertical asymptote.

Quick checklist for sketching rational functions

• Did I factor completely?
• Did I state the domain?
• Did I identify any holes?
• Did I find all intercepts correctly?
• Did I find all vertical asymptotes?
• Did I determine the horizontal or slant asymptote?
• Did I use a sign chart?
• Did I check the graph behavior near each asymptote?

Summary

To sketch a rational function, first factor it completely. Then use the factors to find the domain, holes, x-intercepts, y-intercept, and vertical asymptotes.

Next, determine the end behavior by finding the horizontal or slant asymptote. Finally, use a sign chart and one-sided analysis near vertical asymptotes to decide where each branch goes.

When all these pieces are combined, you can draw an accurate graph even for more complicated rational functions.